OCR C3 2013 June — Question 5 8 marks

Exam BoardOCR
ModuleC3 (Core Mathematics 3)
Year2013
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicModulus function
TypeSolve |linear| > |linear|
DifficultyModerate -0.3 Part (i) is straightforward identification of transformations (horizontal translation and horizontal stretch). Part (ii) requires squaring both sides to eliminate moduli, then solving a quadratic inequality—a standard technique for C3 level. The working is methodical rather than requiring insight, making this slightly easier than average but not trivial due to the algebraic manipulation required.
Spec1.02l Modulus function: notation, relations, equations and inequalities1.02w Graph transformations: simple transformations of f(x)

5
  1. Give full details of a sequence of two transformations needed to transform the graph of \(y = | x |\) to the graph of \(y = | 2 ( x + 3 ) |\).
  2. Solve the inequality \(| x | > | 2 ( x + 3 ) |\), showing all your working.

Question 5:
Part (i):
AnswerMarks Guidance
AnswerMark Guidance
Refer to translation and stretchM1 In either order; ignore details here; allow any equiv wording (such as move or shift for translation) to describe geometrical transformation but not statements such as add 3 to \(x\)
Either: State translation in negative \(x\)-direction by 3A1 Or state translation by \(\begin{pmatrix}-3\\0\end{pmatrix}\); accept horizontal to indicate direction; term 'translate' or 'translation' needed for award of A1
State stretch by factor 2 in \(y\)-directionA1 Or parallel to \(y\)-axis or vertically; term 'stretch' needed for award of A1; these two transformations can be given in either order. SC: if M0 but details of one transformation correct, award B1 for 1/3 (in Either, Or 1, Or 2 cases)
Or 1: State stretch by factor \(\frac{1}{2}\) in \(x\)-directionA1 Or parallel to \(x\)-axis; term 'stretch' needed for award of A1
State translation in negative \(x\)-direction by 3A1 Or state translation by \(\begin{pmatrix}-3\\0\end{pmatrix}\); term 'translate' or 'translation' needed; these two transformations must be in this order – if details correct for M1A1A1 but order wrong, award M1A1A0
Or 2: State translation in negative \(x\)-direction by 6A1 Or state translation by \(\begin{pmatrix}-6\\0\end{pmatrix}\); term 'translate' or 'translation' needed
State stretch by factor \(\frac{1}{2}\) in \(x\)-directionA1 Or parallel to \(x\)-axis; term 'stretch' needed for award of A1; these two transformations must be in this order – if details correct for M1A1A1 but order wrong, award M1A1A0
Part (ii):
AnswerMarks Guidance
AnswerMark Guidance
Either: Solve linear eqn/ineq to obtain critical value \(-6\)B1
Attempt solution of linear eqn/ineq where signs of \(x\) and \(2x\) are differentM1
Obtain critical value \(-2\)A1
Attempt solution of inequalityM1 Using table, sketch, …; implied by correct answer or answer of form \(ab\) (where \(a
Obtain \(-6 < x < -2\)A1 As final answer; must be \(<\) not \(\leq\); allow "\(x>-6\) and \(x<-2\)"
Or: Square both sides to obtain \(x^2 > 4(x^2+6x+9)\)B1 Or equiv
Attempt solution of 3-term quadratic eqn/ineqM1 With same guidelines as in Q2(ii) for factorising and formula
Obtain critical values \(-6\) and \(-2\)A1
Attempt solution of inequalityM1 Using table, sketch, …; implied by correct answer or answer of form \(ab\) (where \(a
Obtain \(-6 < x < -2\)A1 As final answer; must be \(<\) not \(\leq\); allow "\(x>-6\) and \(x<-2\)"
## Question 5:

### Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Refer to translation and stretch | M1 | In either order; ignore details here; allow any equiv wording (such as move or shift for translation) to describe geometrical transformation but not statements such as add 3 to $x$ |
| **Either:** State translation in negative $x$-direction by 3 | A1 | Or state translation by $\begin{pmatrix}-3\\0\end{pmatrix}$; accept horizontal to indicate direction; term 'translate' or 'translation' needed for award of A1 |
| State stretch by factor 2 in $y$-direction | A1 | Or parallel to $y$-axis or vertically; term 'stretch' needed for award of A1; these two transformations can be given in either order. SC: if M0 but details of one transformation correct, award B1 for 1/3 (in Either, Or 1, Or 2 cases) |
| **Or 1:** State stretch by factor $\frac{1}{2}$ in $x$-direction | A1 | Or parallel to $x$-axis; term 'stretch' needed for award of A1 |
| State translation in negative $x$-direction by 3 | A1 | Or state translation by $\begin{pmatrix}-3\\0\end{pmatrix}$; term 'translate' or 'translation' needed; these two transformations must be in this order – if details correct for M1A1A1 but order wrong, award M1A1A0 |
| **Or 2:** State translation in negative $x$-direction by 6 | A1 | Or state translation by $\begin{pmatrix}-6\\0\end{pmatrix}$; term 'translate' or 'translation' needed |
| State stretch by factor $\frac{1}{2}$ in $x$-direction | A1 | Or parallel to $x$-axis; term 'stretch' needed for award of A1; these two transformations must be in this order – if details correct for M1A1A1 but order wrong, award M1A1A0 |

### Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| **Either:** Solve linear eqn/ineq to obtain critical value $-6$ | B1 | |
| Attempt solution of linear eqn/ineq where signs of $x$ and $2x$ are different | M1 | |
| Obtain critical value $-2$ | A1 | |
| Attempt solution of inequality | M1 | Using table, sketch, …; implied by correct answer or answer of form $a<x<b$ or of form $x<a, x>b$ (where $a<b$); allow $\leq$ here |
| Obtain $-6 < x < -2$ | A1 | As final answer; must be $<$ not $\leq$; allow "$x>-6$ and $x<-2$" |
| **Or:** Square both sides to obtain $x^2 > 4(x^2+6x+9)$ | B1 | Or equiv |
| Attempt solution of 3-term quadratic eqn/ineq | M1 | With same guidelines as in Q2(ii) for factorising and formula |
| Obtain critical values $-6$ and $-2$ | A1 | |
| Attempt solution of inequality | M1 | Using table, sketch, …; implied by correct answer or answer of form $a<x<b$ or of form $x<a, x>b$ (where $a<b$); allow $\leq$ here |
| Obtain $-6 < x < -2$ | A1 | As final answer; must be $<$ not $\leq$; allow "$x>-6$ and $x<-2$" |

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5 (i) Give full details of a sequence of two transformations needed to transform the graph of $y = | x |$ to the graph of $y = | 2 ( x + 3 ) |$.\\
(ii) Solve the inequality $| x | > | 2 ( x + 3 ) |$, showing all your working.

\hfill \mbox{\textit{OCR C3 2013 Q5 [8]}}